// the loop routine runs over and over again forever:void loop() { delay(100); digitalWrite(ledI, HIGH); // turn the LED on (HIGH is the voltage level) delay(1000); // wait for a second digitalWrite(ledI, LOW); // turn the LED off by making the voltage LOW delay(3000); // wait for a second digitalWrite(ledL, HIGH); // turn the LED on (HIGH is the voltage level) delay(1000); // wait for a second digitalWrite(ledL, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledO, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledO, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledV, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledV, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledE, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledE, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledY, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledY, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledo, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledo, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledU, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledU, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second}

I then wanted the led I, L, O, V, E, Y, o, U to write high and tried to use digitalWrite but it errored on me. What should I use in its place?

// the loop routine runs over and over again forever:void loop() { delay(100); digitalWrite(ledI, HIGH); // turn the LED on (HIGH is the voltage level) delay(1000); // wait for a second digitalWrite(ledI, LOW); // turn the LED off by making the voltage LOW delay(3000); // wait for a second digitalWrite(ledL, HIGH); // turn the LED on (HIGH is the voltage level) delay(1000); // wait for a second digitalWrite(ledL, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledO, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledO, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledV, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledV, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledE, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledE, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledY, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledY, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledo, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledo, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledU, HIGH); // turn the LED on (HIGH is the voltage level) delay(3000); // wait for a second digitalWrite(ledU, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledI, ledL, ledO, ledV, ledE, ledY, ledo, ledU, HIGH); // turn the LED off by making the voltage LOW delay(1000); // wait for a second digitalWrite(ledI, ledL, ledO, ledV, ledE, ledY, ledo, ledU, LOW); // turn the LED off by making the voltage LOW delay(1000); // wait for a second}

Patience.....

Retroplayer

Ahhh... lol. Yeah, you can't do that. Just put each one on it's own line without any delays in between. It will happen so fast, that it will appear to all turn on at once. The human eye couldn't capture it.

Also, if they are all using the 8 bits (you have 8 letters) of one port, you could just do this:

PORTn = 0xFF and that would turn on all bits at once. n = whatever port they are on. PORTD is pins 0-7. But you will lose your serial port. If you aren't using it, no big deal. It's the only full 8-bit port available.

However, my first suggestion would happen so fast, the results will look the same.

I deleted all delays and it blinked for a tiny fraction of a second but was almost unnoticeable so I deleted the delay after the low and it blinks left to right really fast which is not what I want. Thoughts?

// the loop routine runs over and over again forever:void loop() { delay(1000); digitalWrite(ledI, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledI, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledL, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledL, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledO, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledO, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledV, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledV, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledE, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledE, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledY, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledY, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledo, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledo, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledU, HIGH); // turn the LED on (HIGH is the voltage level) delay(500); // wait for a second digitalWrite(ledU, LOW); // turn the LED off by making the voltage LOW delay(500); // wait for a second digitalWrite(ledI, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledI, LOW); // turn the LED off by making the voltage LOW

digitalWrite(ledL, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledL, LOW); // turn the LED off by making the voltage LOW

digitalWrite(ledO, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledO, LOW); // turn the LED off by making the voltage LOW

digitalWrite(ledV, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledV, LOW); // turn the LED off by making the voltage LOW

digitalWrite(ledE, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledE, LOW); // turn the LED off by making the voltage LOW

digitalWrite(ledY, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledY, LOW); // turn the LED off by making the voltage LOW

digitalWrite(ledo, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledo, LOW); // turn the LED off by making the voltage LOW

digitalWrite(ledU, HIGH); // turn the LED on (HIGH is the voltage level) delay(50); digitalWrite(ledU, LOW); // turn the LED off by making the voltage LOW delay(1000);}

Remember that the Arduino pins are only rated for 40mA max. If you are trying to drive a number of LEDs in parallel then your poor Arduino just can't keep up. If you are driving them in series, then the voltage across the string would have to be much larger than 5V.

If, indeed, the LEDs are in parallel be sure that each LED has its own individual current-limiting resistor and place an NPN transistor from ground to the cathodes of the LEDs and then place the current-limiting resistors from the anode to +5V. Use a 10k resistor from the each Arduino pin to the base of each transistor.

With 330 ohm resistors, each LED will draw about 10mA. So a 2n2222 should be fine. It has a max current of 200mA so it would drive 20 LEDs.

MaJiG

If you move each wire from an Arduino pin one breadboard row to the left and the move the near end of the resistor over to the row with the wire, you'll be gold. Of course the left-most will need a custom swap, but you get the idea...